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Bohr Quantium Condition

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A particle of mass m is attracted to the origin by a force proportional to the distance from the origin F = -kr. Assume that it moves in circular orbit of radius r.

a) Use Newton's law to show that its kinetic energy 1/2mv^2 equals its potential energy 1/2kr^2.

b) Use Bohr quantum condition on the angular momentum (L = nh) to show that the allowed orbitals are given by r^2 - nh*sqrt(km)

c) Find the allowed energies and show that they can be written E_n = nhf where f is the frequency of revolution.

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Solution Summary

The solution provides step-by-step workings, interspersed with explanatory notes, to showing that the various energies and angular momentums of the described particle can be written a certain way.

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a.)
Because, force
F = -k*r = - m*v^2/r = -m*r*w^2 (centripetal force) ......(1)
where, w is angular velocity (rad/s)
=> m*v^2 = k*r^2
=> K.E. = (1/2)*m*v^2 = (1/2)*k*r^2 --Proved

b.)
Because, ...

Solution provided by:
Amrit Lal Ahuja, PhD (IP)
Education
  • BEng, Allahabad University, India
  • MSc , Pune University, India
  • PhD (IP), Pune University, India
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